Question

设x1，x2是方程x^2-x-4=0的两个实数根，求代数式(x1)^3+5(x2)^2+10的值

Answer 1

x1，x2是方程x^2-x-4=0的两个实数根
∴x1+x2=1
x1²-x1-4=0,x2²-x2-4=0
∴x1²=x1+4, x2²=x2+4
(x1)^3+5(x2)^2+10
=x1²·x1+5x2²+10
=(x1+4)x1+5(x2+4)+10
=x1²+4x1+5x2+20+10
=x1+4+4x1+5x2+30
=5(x1+x2)+34
=5×1+34
=39

Answer 2

x1，x2是方程x^2+x-4=0
∴x1²+x1-4=0 x1²=-x1+4
x1³=-x1²+4x1=5x1-4
同理x2²=-x2+4
∴x1³-5x2²+10
=5x1-4+5x2-20+10
=5(x1+x2)-14
=-19

追问

x1³=-x1²+4x1=5x1-4
不理解

追答

∴x1³+5x2²+10
=5x1-4+5(-x2+4)+10
=5(x1-x2)+26
x1+x2=-(-1)/1=1
x1*x2=-4
(x1-x2)²=(x1+x2)²-4x1x2=1+16=17
x1-x2=±√17
=±5√17+26